\(a,D=R\backslash\left\{7\right\}\Rightarrow x1,x2\in D\left(x1\ne x2\right)\)
\(\Rightarrow I=\dfrac{f\left(x1\right)-f\left(x2\right)}{x1-x2}=\dfrac{\dfrac{x1}{x1-7}-\dfrac{x2}{x2-7}}{x1-x2}=\dfrac{\dfrac{-7\left(x1-x2\right)}{\left(x1-7\right)\left(x2-7\right)}}{x1-x2}=\dfrac{-7}{\left(x1-7\right)\left(x2-7\right)}\)
\(trên\left(\text{-∞};7\right)\Rightarrow\left(x1-7\right)\left(x2-7\right)< 0\Rightarrow\dfrac{-7}{\left(x1-7\right)\left(x2-7\right)}< 0\Rightarrow I< 0\Rightarrow nghịch-biến\)
\(b,\Rightarrow D=R\backslash\left\{-1\right\}\Rightarrow x1,x2\in D,x1\ne x2\Rightarrow I=\dfrac{f\left(x1\right)-f\left(x2\right)}{x1-x2}=....\)
\(làm-tương-tự\left(a\right)nếu\left[{}\begin{matrix}I>0\left(đồng-biến\right)\\I< 0\left(nghịch-biến\right)\\I=0\left(ko-đổi\right)\end{matrix}\right.\)
\(c5:\) \(I,J-trung-điểm-AC,BD\Rightarrow\left\{{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IC}=\overrightarrow{0}\\\overrightarrow{IB}+\overrightarrow{ID}=2\overrightarrow{IJ}\end{matrix}\right.\)
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{IB}-\overrightarrow{IA}+\overrightarrow{ID}-\overrightarrow{IC}=\overrightarrow{IB}+\overrightarrow{ID}-\left(\overrightarrow{IA}+\overrightarrow{IC}\right)=\overrightarrow{IB}+\overrightarrow{ID}-\overrightarrow{0}=2\overrightarrow{IJ}\)
